Eigen values ....again :(

Hi,

Here is the question that I am stuck on

$\displaystyle B = \left[\begin{array}{cc}1&1\\1&2\end{array}\right]$

Construct the matrix X that reduces B to diagonal form through the transformation

$\displaystyle D = X^{-1}BX.$ Confirm that D is diagonal, that the diagonal elements of D are the

eigenvalues of B and that det(B) = det(D).

So

$\displaystyle det(B - \lambda I) = \left[\begin{array}{cc}1 - \lambda &1\\1&2 - \lambda \end{array}\right]$

I then get to the point where $\displaystyle \lambda^{2} - 3\lambda + 1$ and therefore my eigen values are $\displaystyle \lambda = \frac{3 \pm \sqrt{5}}{2}$ and my eigen vectors as $\displaystyle \left[\begin{array}{c}-\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right]$ and $\displaystyle \left[\begin{array}{c}\frac{(1 + \sqrt{5})}{2}\\1\end{array}\right]$.

The matrix X is made from the eigen values of B. So $\displaystyle det(X) =\left[\begin{array}{cc}-\frac{(1 + \sqrt{5})}{2}&\frac{(1 + \sqrt{5})}{2}\\1&1\end{array}\right]$. I get as being $\displaystyle -1 -\sqrt{5} $

Now I have to find the cofactor of X and then transpose of the cofactor in order to eventually get the inverse of X.

$\displaystyle X_{cof} = \left[\begin{array}{cc}1&-1\\ \frac{(-1 -\sqrt{5})}{2}&\frac{-1 -\sqrt{5})}{2}

\end{array}\right]$

$\displaystyle X_{cof}T = \left[\begin{array}{cc}1&\frac{-1 -\sqrt{5})}{2}\\ -1&\frac{-1 -\sqrt{5})}{2}

\end{array}\right]$

$\displaystyle X^{-1} = \frac{1}{det(X)} \cdot X_{cof}T $ which I get as $\displaystyle \left[\begin{array}{cc}\frac{1 -\sqrt{5}}{4}&\frac{1}{2}\\ \frac{\sqrt{5} - 1}{4}&\frac{1}{2}\end{array}\right] $

I now have all the components required to find D.

So when I multiply them all together i get $\displaystyle \left[\begin{array}{cc}0&\frac{\sqrt{5} - 1}{2}+1\\ 0&1 +\sqrt{5}\end{array}\right]$

Which is evidently wrong. Can anyone see roughly where i could have gone wrong and then i will post a more detailed process of what i was doing at that stage.

Thanks for reading and any help that come with it